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Math Help - Cartesian Equation of a Plane

  1. #1
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    Exclamation Cartesian Equation of a Plane

    I'm having trouble solving this. I seem to be totally lost. If someone could just tell me the steps I need to take, I'm pretty sure I can work backwards to understand it.

    Question: Determine the Cartestian equation of the plane that passes through the points (1, 4, 5) and (3, 2, 1) and is perpendicular to the plane 2x-y+z-1=0.

    The answer is 3x+5y-1z-18=0


    Thanks
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  2. #2
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    Quote Originally Posted by MATHDUDE2 View Post
    I'm having trouble solving this. I seem to be totally lost. If someone could just tell me the steps I need to take, I'm pretty sure I can work backwards to understand it.

    Question: Determine the Cartestian equation of the plane that passes through the points (1, 4, 5) and (3, 2, 1) and is perpendicular to the plane 2x-y+z-1=0.

    The answer is 3x+5y-1z-18=0


    Thanks
    1. The points P(1,4,5) and Q(3,2,1) are determined by the stationary vectors \vec p and \vec q.

    2. The vector \vec v = \vec q - \vec p and the normal vector of the given plane \vec n span the new plane. Additionally the new plane has to pass through P (or Q).

    3. The normal vector of the new plane is \vec s = \vec v \times \vec n = (-6, -10, 2)

    4. \vec s \cdot ((x,y,z)-\vec p)=0 yields the equation of the new plane.

    EDIT: I've attached the graphs of the 2 planes. The Points P and Q are drawn in blue and red respectively.
    Attached Thumbnails Attached Thumbnails Cartesian Equation of a Plane-senkr_ebenen2pkte.png  
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  3. #3
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    thanks!!
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